Controls
Scene
Readings
- Gravitational force
- 3.34 × 10⁻⁷N
- Force if distance doubles
- 8.34 × 10⁻⁸N
- Force if distance halves
- 1.33 × 10⁻⁶N
- Acceleration of body 1
- 3.34 × 10⁻⁹m/s²
- Acceleration of body 2
- 1.67 × 10⁻⁹m/s²
How to use this simulation
- Start in the Two masses scene with the defaults: 100 kg and 200 kg, 2.0 m apart. The force reads 3.34 × 10⁻⁷ N, a number so small that you could never feel it.
- Drag the distance slider and watch the red marker slide down the graph. Doubling the distance cuts the force to a quarter; the readouts list both cases for you.
- Look at the two acceleration readouts. The force on both bodies is the same, but the lighter body accelerates more. Make the masses equal and the two accelerations match.
- Switch to g at height and depth. Slide from the right (space) to r/R = 1 (the surface) and on into the Earth. The curve falls like 1/r² outside and like a straight line inside.
- Open Satellite orbit and start at 400 km. Read the speed and the period, then raise the altitude. The satellite gets slower and the lap takes longer, even though it has farther to go.
- Push the altitude to the dashed green ring. At about 35,764 km the period matches one day, so the satellite hangs over one spot on the ground.
The apple, the Moon and one idea
The story says that Newton watched an apple fall and asked a strange question: does the pull that brings the apple down reach all the way to the Moon? If it does, then the Moon is also falling. It just keeps missing the Earth, because it is moving sideways so fast.
That one thought joined the heavens and the ground. The same law that makes a cricket ball come down also holds the Moon in its monthly circle, keeps the planets around the Sun, and decides how fast a satellite must fly to stay up.
In exams this topic shows up in three families of questions. First, find the force between two masses or what happens when the distance changes. Second, find g at some height or depth. Third, find the speed or period of a satellite. Each of them is one idea with different clothes, and the simulation lets you try all three.
Newton’s law of gravitation
Every particle attracts every other particle. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centres. It acts along the line joining them.
Put numbers on it and you get F = G m₁ m₂ / r². The constant G is the same everywhere in the universe, G = 6.674 × 10⁻¹¹ N m² kg⁻². It is tiny, which is why you do not feel a pull toward the person sitting next to you, but huge masses like planets make the product large enough to matter.
Two things are worth noticing. The force is always attractive; there is no gravitational push. And the force on body 1 and the force on body 2 are equal and opposite, which is Newton’s third law at work.
F = G m₁ m₂ / r²Attractive force between two point masses
G = 6.674 × 10⁻¹¹ N m² kg⁻²Universal constant of gravitation
F ∝ 1 / r²Inverse-square law: double r, the force is a quarter
How big is the pull between everyday things?
Take two people, 60 kg and 70 kg, standing 1 m apart. Their mutual force is 2.80 × 10⁻⁷ N. That is the weight of a few grains of sand. Friction with the floor beats it by a factor of millions, so nobody drifts together.
Now use the simulation’s defaults: 100 kg and 200 kg at 2.0 m. The force is 3.34 × 10⁻⁷ N. Move the distance to half of that and the force becomes 1.33 × 10⁻⁶ N, four times larger. Go to double and it falls to 8.34 × 10⁻⁸ N.
Gravity is the weakest of the four fundamental forces, yet it rules the sky. Why? Because it never cancels. Electric forces have positive and negative charges that cancel out; the mass of a planet only ever adds up.
The inverse-square law on a graph
Why a square? Imagine the pull spreading out in all directions like light from a bulb. At twice the distance it must cover a sphere with four times the area, so each patch gets a quarter. This picture works for light, sound and gravity alike.
The table fixes the masses at 100 kg and 200 kg and moves them apart. The last column divides each force by the force at 1 m, so you can see the 1, 1/4, 1/9, 1/16, 1/25 pattern with your own eyes.
| Distance r (m) | Force F (N) | F ÷ F at 1 m |
|---|---|---|
| 1 | 1.33 × 10⁻⁶ | 1.000 |
| 2 | 3.34 × 10⁻⁷ | 0.250 |
| 3 | 1.48 × 10⁻⁷ | 0.111 |
| 4 | 8.34 × 10⁻⁸ | 0.063 |
| 5 | 5.34 × 10⁻⁸ | 0.040 |
Equal forces, unequal accelerations
Newton’s third law says the Earth pulls you with exactly the force you pull the Earth. So why does only the apple move? Because acceleration is force divided by mass, and the Earth’s mass is enormous.
The two acceleration readouts show it. With 100 kg and 200 kg, the lighter body accelerates at 3.34 × 10⁻⁹ m/s² and the heavier one at 1.67 × 10⁻⁹ m/s². Same force, different mass, so the lighter body wins the acceleration contest by exactly the ratio of the masses.
For the Earth and an apple, that ratio is about 10²⁵. The Earth does accelerate toward the apple, but by an amount no instrument could ever detect.
Where g comes from
A falling body of mass m feels F = G M m / R² near the Earth’s surface. Newton’s second law says F = m g. The m cancels, and g = G M / R². That is why every object falls with the same acceleration, whatever its mass.
With g = 9.8 m/s² and R = 6,371 km we can even weigh the Earth. M = g R² / G = 5.96 × 10²⁴ kg. Nobody can put the Earth on a scale, but a small pendulum and the constant G are enough. This is why Cavendish’s measurement of G was called weighing the world.
The value of g depends only on the planet’s mass and radius, not on the falling object. A planet with 2 times the Earth’s mass and 2 times its radius has g = 0.50 × 9.8 = 4.90 m/s².
g = G M / R²Acceleration due to gravity at the surface
M = g R² / GMass of the Earth from g
g ∝ M / R²Compare two planets using ratios
Climbing up: g with height
At a height h above the surface the distance from the centre is R + h, so g(h) = G M / (R + h)². Divide by the surface value and you get g(h) = g (R / (R + h))². It drops, but gently at first.
The table shows the numbers. At the height of the International Space Station, 400 km, g is still 8.68 m/s², which is 88.5% of the surface value. A 60 kg person would weigh 521 N there, compared with 588 N on the ground. Astronauts feel weightless not because gravity is gone, but because they are in free fall together with their spacecraft.
At h = R (6,371 km up, so the distance from the centre has doubled) g is a quarter, 2.45 m/s². Far away it keeps shrinking but never quite reaches zero.
g(h) = g · R² / (R + h)²Exact formula above the surface
g(h) ≈ g (1 − 2h / R)Approximation for h much smaller than R
| Height h (km) | g (m/s²) | Share of surface g (%) |
|---|---|---|
| 0 | 9.80 | 100.0 |
| 100 | 9.50 | 96.9 |
| 400 | 8.68 | 88.5 |
| 1,000 | 7.32 | 74.7 |
| 6,371 | 2.45 | 25.0 |
| 35,786 | 0.22 | 2.3 |
Digging down: g with depth
Going below the surface the story changes. For a uniform Earth, only the mass inside your radius pulls you, and the shell above you cancels out. The inside mass shrinks as r³ while the distance shrinks as r², so g falls in proportion to r: g(d) = g (1 − d / R).
Halfway to the centre, g is half of its surface value, 4.90 m/s². At the centre it is exactly zero: the pull comes equally from every side.
The real Earth is denser at the core, so the real g rises a little before it falls. Textbooks use the uniform model, and so does the simulation. Look at the graph: from the left edge, g climbs along a straight line up to r/R = 1, then turns and falls like 1/r². The peak is right at the surface.
g(d) = g (1 − d / R)Below the surface, uniform Earth
g = 0 at the centreEqual pull from every direction
| Depth d (km) | g (m/s²) | Share of surface g (%) |
|---|---|---|
| 0 | 9.80 | 100 |
| 1,593 | 7.35 | 75 |
| 3,186 | 4.90 | 50 |
| 4,778 | 2.45 | 25 |
| 6,371 | 0.00 | 0 |
A satellite is a ball that keeps missing
Throw a stone sideways from a mountain. Throw harder and it lands farther away. Throw hard enough and the ground curves away as fast as the stone falls, so the stone never lands. That stone is in orbit.
The gravitational pull supplies the centripetal force: m v² / r = G M m / r². The mass m of the satellite cancels, so the speed depends only on the orbit radius r = R + h: v = √(G M / r) = √(g R² / (R + h)).
The period is the circumference divided by the speed. A satellite at 400 km flies at 7.66 km/s, about 27,593 km/h, and goes round once in 92.5 minutes. That is 15.6 sunrises every day for the crew.
Higher means slower. That feels backwards, because a higher satellite has more energy. But it is moving through a weaker field, so a smaller speed is enough to hold it, and it also has a longer path to cover. The result is that the period grows quickly with altitude.
v = √(G M / (R + h))Orbital speed; independent of the satellite’s mass
T = 2π (R + h) / vTime for one lap
T² ∝ (R + h)³Kepler’s third law
| Altitude h (km) | Orbital speed (km/s) | Time period (min) |
|---|---|---|
| 200 | 7.78 | 88.4 |
| 400 | 7.66 | 92.5 |
| 2,000 | 6.89 | 127.2 |
| 20,200 | 3.87 | 719.2 |
| 35,786 | 3.07 | 1,437.2 |
The geostationary orbit
A satellite with a period of one sidereal day, over the equator and moving the same way as the Earth, stays above one spot. That makes it perfect for television and weather pictures: a dish on the roof never has to move.
Setting T = 24 hours in Kepler’s law fixes the radius. The altitude comes out as 35,764 km, the speed 3.07 km/s and the period 23.93 hours. You cannot choose a different height; one day fixes the orbit. In the simulation this is the dashed green ring.
Doubling the orbit radius multiplies the period by 2.83, not by 2. That is the 3/2 power in action, since 2 raised to 1.5 is 2.83.
Escape speed and weightlessness
To leave the Earth’s pull for good a body needs the escape speed, v = √(2 g R). From the surface that is 11.17 km/s. A satellite skimming the surface would need only 7.90 km/s, which is the same figure divided by √2. The readout in the Satellite orbit scene shows the escape speed from wherever the satellite is.
Escape does not require going that fast all the time. A rocket can climb slowly if it keeps burning fuel. The number is the speed needed if the engines switch off at the start and the body is thrown.
Weightlessness in orbit has nothing to do with escaping gravity. Everything inside the station falls at the same rate as the station itself, so nothing presses on anything else. A scale under your feet reads zero because the scale is falling too.
v(escape) = √(2 g R) = √2 · v(orbit at the surface)Energy needed to reach infinity
Solved problems
Five problems in the style of the exam, solved with the same functions that drive the animation.
Problem 1: change the distance
Two bodies attract each other with 3.34 × 10⁻⁷ N at 2.0 m. What is the force if the distance is made three times as large?
F ∝ 1/r², so tripling r divides the force by 9: F = 3.34 × 10⁻⁷ ÷ 9. The table shows the same pattern from 1 m to 3 m, where the ratio drops from 1 to 0.111.
Problem 2: the Moon’s acceleration
The Moon is 3.84 × 10⁸ m from the Earth, and the masses are 5.97 × 10²⁴ kg and 7.35 × 10²² kg. Find the force between them and the Moon’s acceleration.
F = G M m / r² = 1.99 × 10²⁰ N. The Moon’s acceleration is F / m = 2.70 × 10⁻³ m/s². Newton’s great check: the Moon is 60.3 Earth radii away, so by the inverse-square law its acceleration should be g divided by 60.3² = 2.70 × 10⁻³ m/s². The two answers agree.
Problem 3: g on another planet
A planet has 2 times the Earth’s mass and 2 times its radius. What is g on it?
g ∝ M / R², so the ratio is 2 ÷ 2² = 0.50. g = 0.50 × 9.8 = 4.90 m/s². More mass, but a much larger radius, so the surface is weaker than ours.
Problem 4: a satellite at 400 km
Find the orbital speed and period of a satellite 400 km above the Earth. Use g = 9.8 m/s² and R = 6,371 km.
r = R + h = 6,371 + 400 km. v = √(g R² / r) = 7.66 km/s. T = 2π r / v = 92.5 min. Notice that the satellite’s mass was never needed.
Problem 5: a geostationary satellite
At what height does a satellite appear fixed over one point of the equator?
Its period must be one sidereal day. Put T = 2π r / v with v = √(g R² / r) and solve for r. The altitude is 35,764 km, with orbital speed 3.07 km/s.
Mistakes that cost marks
Check this list before an exam. Each item has taken marks from someone.
- Measuring r from the surface instead of the centre. A satellite at height h has r = R + h.
- Using g = 9.8 m/s² at height or depth. g changes; use the formulas.
- Writing g(d) = g (R − d)² / R². That is the height formula; below the surface g is linear in distance from the centre.
- Thinking a heavier satellite orbits faster. Mass cancels; speed depends only on the orbit radius.
- Saying astronauts are weightless because there is no gravity. There is plenty; they are in free fall.
- Forgetting the units: G is in N m² kg⁻², kilometres must be converted to metres before using the formulas.
- Calling the escape speed 11.2 m/s. It is 11.2 km/s, which is about 40,000 km/h.
Where else you meet it
Ocean tides rise because the Moon’s pull is slightly stronger on the near side of the Earth than the far side. GPS satellites sit at about 20,200 km, where the period is half a day, so each receiver sees several at once. Weather satellites in low orbit pass over the poles to scan the whole globe.
The same law, applied to the Sun and the planets, gives Kepler’s three laws for free. If you want to see the shape of an elliptical orbit instead of a circle, try the Kepler orbits animation next.
Frequently asked questions
What does Newton’s law of gravitation say?
Every two masses attract each other with a force proportional to the product of the masses and inversely proportional to the square of the distance between their centres: F = G m₁ m₂ / r².
What is the value of G and does it change?
G = 6.674 × 10⁻¹¹ N m² kg⁻². It is a universal constant and does not change from place to place. The value of g, on the other hand, changes with height, depth and from planet to planet.
How does g change with height and with depth?
Above the surface g falls as 1/(R + h)². Below the surface, for a uniform Earth, g falls linearly with depth and is zero at the centre. It is greatest at the surface.
How do you find the orbital speed of a satellite?
Use v = √(G M / r) = √(g R² / (R + h)). The speed depends only on the orbit radius, not on the mass of the satellite.
Why do astronauts float in a space station?
Gravity at that height is still about ninety percent of its value on the ground. The astronauts float because they and the station are in free fall together, so there is no contact force between them.
What is a geostationary satellite?
A satellite over the equator with a period of one day, so it stays above one point on the ground. Its height is fixed at about 36,000 km above the surface.
What is escape speed?
The least speed a body needs at the surface to move away from the Earth forever without further thrust. It equals √(2 g R), about 11.2 km/s for the Earth.
Keep studying this topic
The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.
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HSC Physics: syllabus and preparation
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Composition: The Uses of Electricity
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Paragraph: Science in Everyday Life
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Natural selection: peppered moths and the sooty tree
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Immune response: primary and secondary antibody response
Biology
Pollination and fertilisation
Mathematics
Pythagorean Theorem
Mathematics
Trigonometric Ratios and the Unit Circle
Mathematics
Quadratic functions and their graphs
Mathematics
Circle theorems
Mathematics
Heights and distances: elevation and depression angles
Mathematics
Probability: experimental vs theoretical probability
Mathematics
Arithmetic and Geometric Sequences
Mathematics
Mean, Median and Mode
Mathematics
Sets and Venn Diagrams
Mathematics
Linear equations: graphs, slope and simultaneous pairs
Mathematics
Graphs of sin, cos and tan
Mathematics
Completing the Square and the Quadratic Formula
Mathematics
Area of a Circle: Proof, Formula and Solved Problems
Mathematics
Volume and Surface Area of Solids: Cube, Cuboid, Cylinder, Cone, Sphere
Mathematics
Similar Triangles: Scale Factor, AA, SSS, SAS and Shadow Heights
Mathematics
Coordinate Geometry: Distance, Midpoint, Slope, Section Formula and Locus
Mathematics
Conic Sections: Circle, Ellipse, Parabola and Hyperbola
Mathematics
Function Transformations: Shifts, Stretches and Reflections
Mathematics
Derivative as the Slope of a Tangent: Secant to Tangent
Mathematics
Riemann Sums and Integration: Area Under a Curve
Mathematics
Limits and the ε–δ Definition: A Visual Guide
Mathematics
3D Vectors: Sum, Dot Product and Cross Product
Mathematics
Matrix Transformations: Rotation, Reflection, Shear and the Determinant
Mathematics
Complex Numbers: Sum, Product, Powers and Roots on the Argand Plane
Mathematics






















